Question

Show that the origin is equidistant from the lines $4x+3y+10=0;5x-12y+26=0$ and $7x+24y=50$.

Answer 1

4x+3y+10=0 ---- (1)

5x-12y+26=0 ----- (2)

7x+24y=50 ------ (3)

The distance between origin & line (1) = $|\frac{4\times 0+30+10}{\sqrt{16+9}}|=\frac{10}{5}=2$

The distance between origin & line (2) = $|\frac{5\times 0-12\times 0+26}{\sqrt{25+144}}|=\frac{26}{13}=2$

The distance between origin & line (3) = $|\frac{7\times 0+24\times 0-50}{\sqrt{(24{)}^2+(7{)}^2}}|=\frac{50}{25}=2$

$\therefore$ This distance between origin & all three lines is same.